Is it meaningful to distinguish a one-object category and its opposite?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
As in the title, for a category with only one object $mathcal C$, $mathcal C^op$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^op$ play different roles?
Thank you in advance.
category-theory
asked Jan 28 at 20:01
Display NameDisplay Name
449212
$endgroup$
add a comment |
$begingroup$
As in the title, for a category with only one object $mathcal C$, $mathcal C^op$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^op$ play different roles?
Thank you in advance.
category-theory
asked Jan 28 at 20:01
Display NameDisplay Name
449212
$endgroup$
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
add a comment |
$begingroup$
As in the title, for a category with only one object $mathcal C$, $mathcal C^op$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^op$ play different roles?
Thank you in advance.
category-theory
asked Jan 28 at 20:01
Display NameDisplay Name
449212
$endgroup$
As in the title, for a category with only one object $mathcal C$, $mathcal C^op$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^op$ play different roles?
Thank you in advance.
category-theory
category-theory
asked Jan 28 at 20:01
Display NameDisplay Name
449212
asked Jan 28 at 20:01
Display NameDisplay Name
449212
asked Jan 28 at 20:01
Display NameDisplay Name
449212
asked Jan 28 at 20:01
Display NameDisplay Name
449212
asked Jan 28 at 20:01
Display NameDisplay Name
449212
449212
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
add a comment |
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$mathcalC$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcalC^op$ corresponds to $M^op$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^op$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^op$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,5121437
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add a comment |
$begingroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^mathrmopto Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^mathrmop$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.1k23372
$endgroup$
$begingroup$
Perhaps providing an example where $M$ and $M^op$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$mathcalC$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcalC^op$ corresponds to $M^op$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^op$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^op$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,5121437
$endgroup$
add a comment |
$begingroup$
$mathcalC$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcalC^op$ corresponds to $M^op$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^op$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^op$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,5121437
$endgroup$
add a comment |
$begingroup$
$mathcalC$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcalC^op$ corresponds to $M^op$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^op$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^op$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,5121437
$endgroup$
$mathcalC$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcalC^op$ corresponds to $M^op$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^op$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^op$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,5121437
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,5121437
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,5121437
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,5121437
7,5121437
add a comment |
add a comment |
$begingroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^mathrmopto Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^mathrmop$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.1k23372
$endgroup$
$begingroup$
Perhaps providing an example where $M$ and $M^op$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
add a comment |
$begingroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^mathrmopto Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^mathrmop$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.1k23372
$endgroup$
$begingroup$
Perhaps providing an example where $M$ and $M^op$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
add a comment |
$begingroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^mathrmopto Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^mathrmop$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.1k23372
$endgroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^mathrmopto Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^mathrmop$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.1k23372
edited Jan 28 at 22:16
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.1k23372
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.1k23372
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.1k23372
33.1k23372
$begingroup$
Perhaps providing an example where $M$ and $M^op$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
add a comment |
$begingroup$
Perhaps providing an example where $M$ and $M^op$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
$begingroup$
Perhaps providing an example where $M$ and $M^op$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
$begingroup$
Perhaps providing an example where $M$ and $M^op$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
add a comment |
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$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31